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# Is xy > 0?

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Is xy > 0? [#permalink]

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16 Nov 2009, 14:27
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Is xy > 0?

(1) x - y > -2
(2) x - 2y < -6

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-xy-0-1-x-y-2-2-x-2y-114731.html
[Reveal] Spoiler: OA
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Re: Is xy > 0? [#permalink]

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16 Nov 2009, 14:58
study wrote:
is xy > 0?

1) x - y > -2
2) x - 2y < -6

1) x + 2 > y
If y is +ve, x still could be +ve or -ve.
If x is +ve, y still could be +ve or -ve.

2) x +6 < 2y
x and y could be anything.

1+2 togather:
1) -x - 2 < - y
2) x +6 < 2y

Add up them
-x - 2 < - y
x +6 < 2y
4 < y

We know y is +ve and > 4. x is also +ve.

Suff. C.
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Re: Is xy > 0? [#permalink]

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16 Nov 2009, 15:16
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study wrote:
is xy > 0?

1) x - y > -2
2) x - 2y < -6

Question basically asks whether x and y have the same sign.

(1) y<x+2, can we find the pair of x and y with different signs satisfying the inequality given? Sure x=1 y any negative value. Can we find the pair of x and y with the same sign satisfying the inequality given? Sure x=5 y=1. So not sufficient to conclude whether x and y have the same sign or not.

(2) y>x/2+6. The same here. We can find x and y with different signs, as well as the same sign to satisfy inequality given. Not sufficient.

(1)+(2) Remember we can subtract inequalities with the signs in different direction. (1)-(2) 0<x/2-4 --> 8<x. So we get that x is positive and we know from (2) that y>x/2+6. x/2+6 is some positive value and y is more than that positive value, or in another words y is positive too. Sufficient.

Answer: C.

The problem above can be solved in different way:
(1) y<x+2 represent all x,y points which are below the line y=x+2. If you draw this line you'll see that this are consists of the points which are in all quadrants. Or there are xy points with all combinations of x and y: positive x negative y, positive x positive y etc. and we are looking whether x and y are eother from the I quadrant or from the III. Not sufficient.

(2) y>x/2+6. Again if you draw the line y=x/2+6, the points satisfying the inequality given would be above of this line. The points would be in quadrants I, II, or III. Not sufficient.

(1)+(2) The area you'll receive from the two inequalities y<x+2 and y>x/2+6 (below the first and above the second line) will be consisting from the points which are only I quadrant, which means that x and y are both positive. Hence xy is positive. Sufficient.

Answer: C.
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Re: Is xy > 0? [#permalink]

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16 Nov 2009, 15:24
Bunuel wrote:
study wrote:
is xy > 0?

1) x - y > -2
2) x - 2y < -6

Question basically asks whether x and y have the same sign.

(1) y<x+2, can we find the pair of x and y with different signs satisfying the inequality given? Sure x=1 y any negative value. Can we find the pair of x and y with the same sign satisfying the inequality given? Sure x=5 y=1. So not sufficient to conclude whether x and y have the same sign or not.

(2) y>x/2+6. The same here. We can find x and y with different signs, as well as the same sign to satisfy inequality given. Not sufficient.

(1)+(2) Remember we can subtract inequalities with the signs in different direction. (1)-(2) 0<x/2-4 --> 8<x. So we get that x is positive and we know from (2) that y>x/2+6. x/2+6 is some positive value and y is more than that positive value, or in another words y is positive too. Sufficient.

Answer: C.

The problem above can be solved in different way:
(1) y<x+2 represent all x,y points which are below the line y=x+2. If you draw this line you'll see that this are consists of the points which are in all quadrants. Or there are xy points with all combinations of x and y: positive x negative y, positive x positive y etc. and we are looking whether x and y are eother from the I quadrant or from the III. Not sufficient.

(2) y>x/2+6. Again if you draw the line y=x/2+6, the points satisfying the inequality given would be above of this line. The points would be in quadrants I, II, or III. Not sufficient.

(1)+(2) The area you'll receive from the two inequalities y<x+2 and y>x/2+6 (below the first and above the second line) will be consisting from the points which are only I quadrant, which means that x and y are both positive. Hence xy is positive. Sufficient.

Answer: C.

as a review..subtract with signs in different directions and add with signs in same direction?
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Re: Is xy > 0? [#permalink]

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16 Nov 2009, 15:35
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lagomez wrote:
as a review..subtract with signs in different directions and add with signs in same direction?

Yes.

You can only add inequalities when their signs are in the same direction.

You can only apply subtraction when their signs are in the opposite directions.
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Re: Is xy > 0? [#permalink]

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16 Nov 2009, 21:48
Bunuel wrote:
lagomez wrote:
as a review..subtract with signs in different directions and add with signs in same direction?

Yes.

You can only add inequalities when their signs are in the same direction.

You can only apply subtraction when their signs are in the opposite directions.

Bunuel, can you give examples along to make this more clear?

Which sign would the final inequalities take when you subtract them?

Thanks in advance
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Re: Is xy > 0? [#permalink]

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17 Nov 2009, 05:46
lonewolf wrote:
Bunuel, can you give examples along to make this more clear?

Which sign would the final inequalities take when you subtract them?

Thanks in advance

1. a>b and 2. c<d --> a-c>b-d, so the sign > of the 1. (the sign of the one you subtracting from).

You can look at this in another way: c<d means -c>-d, now add inequalities a-c>b-d.

eg. 5>3 and 2<7 --> 5-2>3-7, 3>-4.

Hope it's clear.
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Re: Is xy > 0? [#permalink]

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17 Nov 2009, 06:09
Bunuel wrote:
lonewolf wrote:
Bunuel, can you give examples along to make this more clear?

Which sign would the final inequalities take when you subtract them?

Thanks in advance

1. a>b and 2. c<d --> a-c>b-d, so the sign > of the 1. (the sign of the one you subtracting from).

You can look at this in another way: c<d means -c>-d, now add inequalities a-c>b-d.

eg. 5>3 and 2<7 --> 5-2>3-7, 3>-4.

Hope it's clear.

Great! Crystal Clear
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Re: Is xy > 0? [#permalink]

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03 Jan 2015, 03:36
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Is xy>0?

1) x-y > -2
2) x-2y < -6

---------------------------------------

Can somebody help me with this?

I can solve that 1 or 2 alone are not sufficient so a) b) and d) are ruled out.

But how to solve two algebraic inequalities together to find out that together these two are sufficient or not?

Thanks
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Re: Is xy > 0? [#permalink]

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03 Jan 2015, 03:54
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xy > 0? means they are asking if x and y have the same sign

STAT1 x -y > -2
means that x > y-2
But we cannot say whether x and y have the same sign using this
So, INSUFFICIENT

STAT2 x-2y < -6
x < 2y - 6
But we cannot say whether x and y have the same sign using this
So, INSUFFICIENT

Cobining both of them

x > y -2 ...(1)
and x < 2y -6
Second equation can be written as
-x > -(2y -6)
-x > 6 - 2y ...(2)

Since inequalities (1) and (2) have the same sign so we can add them
x - x > y - 2 + 6 - 2y
=> 0 > 4-y
or y > 4
and x > y - 2
Since y > 4 so y-2 will be greater than 2
So, x > 2
Means both x and y have the same sign (+ve)
So, xy > 0

So, Answer will be C
Hope it helps!
chittakukkar wrote:
Is xy>0?

1) x-y > -2
2) x-2y < -6

---------------------------------------

Can somebody help me with this?

I can solve that 1 or 2 alone are not sufficient so a) b) and d) are ruled out.

But how to solve two algebraic inequalities together to find out that together these two are sufficient or not?

Thanks

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Re: Is xy > 0? [#permalink]

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03 Jan 2015, 14:27
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Hi chittakukkar,

When you're dealing with multiple inequalities, sometimes the "key" to spotting the hidden relationships/patterns is in HOW you write your inequalities. Since you're comfortable that both Fact 1 and Fact 2 are individually INSUFFICIENT, I won't rehash any of that work. I am going to rewrite the given inequalities in a certain way though...

The question asks if (X)(Y) > 0?

Fact 1: X - Y > -2

X + 2 > Y

Fact 2: X - 2Y < -6

X + 6 < 2Y

With these two inequalities, we can make some additional deductions....

To start...
Y < X + 2

By comparison, we know that (X+2) has to be less than (X+6), so we can add THAT in...

Y < (X + 2) < (X + 6)

And we already knew that (X+6) is less than 2Y, so we can now add THAT in...

Y < (X + 2) < (X + 6) < 2Y

Notice how Y is ultimately less than 2Y....think about MUST be true about that relationship....

Could Y = 0? No....because 0 is NOT less than 2(0).
Could Y = negative? No...because in that scenario 2Y would be less than Y.

So, Y MUST be positive.

Next, we need 2Y to be big enough to account for the difference in (X+2) and (X+6). Since those two terms differ by "+4", 2Y and Y have to differ by MORE than 4.

That does NOT happen when.....
Y = 1, 2Y = 2
Y = 2, 2Y = 4
Y = 3, 2Y = 6
Y = 4, 2Y = 8

It DOES happen when Y > 4 though....
Y = 4.1, 2Y = 8.2

By extension, since X+2 > 4, X MUST ALSO be POSITIVE.

Since X and Y BOTH must be POSITIVE, the answer to the question is ALWAYS YES.
Combined, SUFFICIENT

Final Answer:
[Reveal] Spoiler:
C

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Re: Is xy > 0? [#permalink]

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05 Jan 2015, 05:14
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Is xy>0?

Note that question basically asks whether $$x$$ and $$y$$ have the same sign.

(1) x-y > -2 --> we can have an YES answer, if for example $$x$$ and $$y$$ are both positive ($$x=10$$ and $$y=1$$) as well as a NO answer, if for example $$x$$ is positive and $$y$$ is negative ($$x=10$$ and $$y=-10$$). Not sufficient.

(2) x-2y <-6 --> again it' easy to get an YES answer, if for example $$x$$ and $$y$$ are both positive ($$x=1$$ and $$y=10$$) as well as a NO answer, if for example $$x$$ is negative and $$y$$ is positive ($$x=-1$$ and $$y=10$$). Not sufficient.

You can get that the the two statement individually are not sufficient in another way too: we have (1) $$y<x+2$$ and (2) $$y>\frac{x}{2}+3$$. We are asked whether $$x$$ and $$y$$ have the same sign or whether the points (x,y) are in the I or III quadrant ONLY. But all (x,y) points below the line $$y=x+2$$ (for 1) and all (x, y) points above the line $$y=\frac{x}{2}+3$$ can not lie only I or III quadrant: points above or below some line (not parallel to axis) lie at least in 3 quadrants.

(1)+(2) Now, remember that we can subtract inequalities with the signs in opposite direction --> subtract (2) from (1): $$x-y-(x-2y)>-2-(-6)$$ --> $$y>4$$. As $$y>4$$ and (from 1) $$x>y-2$$ then $$x>2$$ (because we can add inequalities when their signs are in the same direction, so: $$y+x>4+(y-2)$$ --> $$x>2$$) --> we have that $$y>4$$ and $$x>2$$: both $$x$$ and $$y$$ are positive. Sufficient.

Answer: C.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-xy-0-1-x-y-2-2-x-2y-114731.html
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Is xy > 0? [#permalink]

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05 Jan 2015, 05:15
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Graphic approach:

Is xy > 0?

$$xy>0$$ means that x and y must have the same sign, so point (x, y) must be either in the first or the third quadrant (green regions).

(1) x-y > -2 --> $$y<x+2$$ --> the area below blue line ($$y=x+2$$). (x, y) may or may not be in green region. Not sufficient.

(2) x-2y < -6 --> $$y>\frac{x}{2}+3$$ --> the area above red line ($$y>\frac{x}{2}+3$$). (x, y) may or may not be in green region. Not sufficient.

(1)+(2) Below blue line and above red line, is yellow region, which is entirely in I quadrant (where $$y>4$$ and $$x>2$$) --> $$xy>0$$. Sufficient.

Answer: C.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-xy-0-1-x-y-2-2-x-2y-114731.html
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